STM32F0xx UID decoding

Question

In my project, I need to make readable 96-bit UID. In STM32F0xx reference manual RM0091 section 33.1 Unique device ID register, it is described how the number is encoded:

struct Uid {
    uint16_t X;  // x-coordinate
    uint16_t Y;  // y-coordinate
    uint8_t WAF;  // Wafer number
    char LOT[7];  // Lot number
};

My question is how x and y coordinates are encoded, if 0x0000 is in middle of wafer or one side?

Here are some examples which I see in X-coordinate: 0x0000, 0x0001, 0x0006, 0x8001, 0x8004, 0x801f, .... some numbers have the MSB bit set.

In Y-coordinate I always see numbers where MSB is NOT set, like: 0x0001, 0x0003, 0x0005, 0x001f, 0x0030, 0x003f, ...

I don't have so many samples to read many numbers and get better statistics of numbering so my question is: what does it mean if a number has the MSB bit set in the coordinate? I think that this is sign, but I'm not sure.

The reason why I am interested about this, is that I need to decrease the size of this number and remove some bits from this UID. My idea is to remove bits 11-14 from each coordinate to keep position and (probably) sign and make this number smaller and still unique. I need to decrease it by 8 bits only.

MCU is STM32F0xx

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UPDATE: I have small statistic from 75 pcs of STM32F031 and here are unique values for:

X: 0003, 0008, 0009, 000a, 000b, 000e, 0010, 0013, 0014, 0015, 0016, 0018, 001c, 0021, 0024, 002a, 8001, 8002, 801b, 801c, 801d, 8020
Y: 0003, 0005, 0006, 0007, 000c, 000f, 0010, 0011, 0013, 0014, 0016, 0017, 0018, 0019, 001a, 001b, 001e, 001f, 0020, 0022, 0024, 0028, 0029, 002a, 002c, 002d, 002e, 0030, 0032, 0033, 0034, 0035, 0037, 0038, 0039, 003a, 003c, 003f, 0042, 0044, 0045, 0046, 0047, 0048, 0049, 004b, 004e
WAF: 0b, 0c, 0d, 0e, 18

Seems that X and Y and WAFER is not BCD but HEX and MSB bit in X coordinate will be probably be sign.

Answer 1

My question is how x and y coordinates are encoded

You don't actually care how they are encoded. They are just a source of unique bits. You take the whole Uid and hash it and then use the hash. The hash can be trimmed to a length that is reasonable for your expected number of units sold over the lifetime of the device, and for the number of usable bits or nibbles in the Uid.

Since the Uid is 12 bytes long, you don't need to use more than 24 hex digits of the hash. Since you discovered that there are at least 4 nibbles that are unused across the X, Y and WAF fields, so 20 hex characters out of a longer hash will be plenty for collision-free mapping from Uid to a hash. Realistically, given that you're probably not selling billions of those devices, 16 hex characters or 8 bytes of the hash is all you'll need.

If this is for licensing purposes then also remember that hash collisions will be inconsequential. The likelihood that the same customer gets two different units with the same partial hash is so low, that you may as well let them enjoy this as a gesture of good will. It won't ever happen in practice.

Answer 2

If all you need to reduce number of bits, but don't know which ones are significant, you can always "xor" them together. for example, xor upper and lower bytes: 0x801f -> 0x80 ^ 0x1f = 0x9f

This requires that your high and low bytes are independent (for example if numbers have a step of 0x101, then the algorithm would not work well). But from your data, it looks like you will be fine.